how to get the most frequent items

问题:

I am working on an application which has a large array containing lines of numbers,

transNum[20000][200]//this is the 2d array containing the numbers and always keep track of the line numbers

I am using a nested loop to look for the most frequent items. which is

for(int i=0/*,lineitems=0*/;i<lineCounter;i++)
  {
      for(int j=0,shows=1;j<lineitem1[i];j++)
      {
          for(int t=i+1;t<lineCounter;t++)
          {
              for(int s=0;s<lineitem1[t];s++)
              {
                  if(transNum[i][j]==transNum[t][s])
                      shows++;
              }
          }

          if(shows/lineCounter>=0.2)
          {

              freItem[i][lineitem2[i]]=transNum[i][j];
              lineitem2[i]++;
          }
      }

  }

when I was doing tests using small input arrays like test[200][200], this loop works fine and the computing time is acceptable, but when I try to process the array contains 12000 lines, the computing time is too long, so I am thinking if there are other ways to compute the frequent items rather than using this loop.I just ran a test on 10688 lines, and the time to get all the frequent item is 825805ms, which is way to expensive.

回答1:

Depends on your input. If you are also inserting the data in the same code then you can count frequent items as you insert them.

---

Here is a pseudo-C solution:

int counts[1000000];

while(each number as n)
{
    counts[n]++;
    // then insert number into array
}

EDIT #2: Make sure, so you don't get unexpected results, to initialize all the items in the array to zero.

回答2:

Bear in mind this is an O(n^2) algorithm at best and could be worse. That means the number of operations is proportional to the count of the items squared. After a certain number of lines, performance will degrade rapidly and there's nothing you can do about it except to improve the algorithm.

回答3:

The Multiset implementation from Google Guava project might be useful in such cases. You could store items there and then retrieve set of values with count of each occurrence.

回答4:

Gave the algorithm for this one some thought. Here's the solution I came up with:

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Random;

public class NumberTotalizerTest {

    public static void main(String args[]) {

        HashMap<Integer,Integer> hashMap = new HashMap<Integer,Integer>();

        // Number input
        Random randomGenerator = new Random();
        for (int i = 1; i <= 50; ++i ) {
            int randomInt = randomGenerator.nextInt(15);
            System.out.println("Generated : " + randomInt);

            Integer tempInt = hashMap.get(randomInt);

            // Counting takes place here
            hashMap.put(randomInt, tempInt==null?1:(tempInt+1) );
        }

        // Sorting and display
        Iterator itr =  sortByValue(hashMap).iterator();

        System.out.println( "Occurences from lowest to highest:" );

        while(itr.hasNext()){
            int key = (Integer) itr.next();

            System.out.println( "Number: " + key + ", occurences: " + hashMap.get(key));
        }
    }

     public static List sortByValue(final Map m) {
        List keys = new ArrayList();
        keys.addAll(m.keySet());
        Collections.sort(keys, new Comparator() {
            public int compare(Object o1, Object o2) {
                Object v1 = m.get(o1);
                Object v2 = m.get(o2);
                if (v1 == null) {
                    return (v2 == null) ? 0 : 1;
                }
                else if (v1 instanceof Comparable) {
                    return ((Comparable) v1).compareTo(v2);
                }
                else {
                    return 0;
                }
            }
        });
        return keys;
    }
}